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[tex]\int\limits_0^{\pi/2} \sin^{2a+1}(x)\,dx[/tex]

Has a Factorial Form:

[tex] {(2^a a!)}^2 \over (2a+1)![/tex]

What is the process behind going from that integral to that factorial form?

My approach which is not very insightful:

I used mathematica to calculate the integral to return:

[tex]\pmb{\frac{\sqrt{\pi } \text{Gamma}[1+a]}{2 \text{Gamma}\left[\frac{3}{2}+a\right]}}[/tex]

I know Gamma[1+a] = a! and Gamma[3/2+a] has a factorial form also but doesn't help me to reduce to that form.

Griffiths just says that integral equals (2*4*...2a)/(1*3*5...*[2a+1]) to get from this to that factorial form is easy but I got lost in his integration.

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# Relating integral of powers of Sin b/w 0 and pi/2 to factorial form

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